Question

Calculate the amount of heat (kcal) released when 50.0g of steam at 100*c hits the skin, condenses, and cools to a body temperature of 37*c

Answer 1

conversion of steam at 100 C to water at 100 C :

m = mass of the steam = 50 g

L = latent heat of condensation of steam to water = 79.7 J/g

Heat released in conversion of steam to water is given as

Q₁ = mL

inserting the values

Q₁ = (50) (79.7)

Q₁ = 3985 cal

conversion of water at 100 C to water at 37 C :

ΔT = change in temperature = 100 - 37 = 63 C

c = specific heat of water = 1 cal/(gC)

m = mass of water = 50 g

Heat released in change of temperature is given as

Q₂ = m c ΔT

Q₂ = 50 x 1 x 63

Q₂ = 3150 cal

total heat released is given as

Q = Q₁ + Q₂

Q = 3985 + 3150

Q = 7135 cal

Answer 2

As the steam touches the skin, it undergoes a phase change and releases latent heat due to the phase change. As it reaches equilibrium, it releases sensible heat. We calculate as follows:

Q = latent heat + sensible Heat
Q = 2.26 kJ / g (50.0 g) + 50.0 g ( 4.18 J / g C) (37 C - 100 C) ( 1 kJ / 1000 J)
Q = 99.833 kJ